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QPages 419-421: Q16: HC Verma: Concepts of Physics - Volume 2 Semiconductors and Semiconductor Devices Pages 419-421: Q16 Solution: In a p-n junction, the depletion region is 400 nm wide and an…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the barrier height in volts as an energy in joules, forgetting that a conduction electron’s kinetic energy is $E = eV$, which numerically equals the same value in electronvolts (eV). Always multiply the potential difference by the elementary charge when converting to joules, or express the result directly in eV to avoid mixing V and J units.

In a p-n junction, the depletion region is 400 nm wide and an electric field of 5×105 V m−15 \times 10^5 \text{ V m}^{-1} exists in it. a) Find the height of the potential barrier. b) What should be the minimum kinetic energy of a conduction electron which can diffuse from the n-side to the p-side?

Step-by-Step Verified Solution

**Answer:** (a) Potential barrier = 0.2 V (b) Minimum kinetic energy = 0.2 eV. **Step-by-Step Explanation:** Given: Depletion width = 400 nm = 4 _ 10__ m, E = 5 _ 10_ V/m. (a) Potential barrier = E _ d = 5 _ 10_ _ 4 _ 10__ = 0.2 V. (b) An electron diffusing from n-side to p-side must overcome this barrier, so minimum KE = 0.2 eV .

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